July 2018
Integer-antimagic spectra of disjoint unions of
cycles
Wai Chee Shiu
Hong Kong Baptist University, wcshiu@hkbu.edu.hk
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Shiu, Wai Chee (2018) "Integer-antimagic spectra of disjoint unions of cycles,"Theory and Applications of Graphs: Vol. 5 : Iss. 2 , Article 3.
DOI: 10.20429/tag.2018.050203
1
Introduction and some useful lemmas
LetGbe a simple graph. For any nontrivial abelian groupA(written additively), letA∗ = A\{0}, where 0 is the additive identity ofA. Let a mappingf :E(G)→A∗ be an edge label-ing ofGandf+ :V(G)→Abe its induced labeling, which is defined byf+(v) = ∑
uv∈E(G) f(uv).
If there exists an edge labeling f whose induced labeling f+ on V(G) is injective, then we
say that f is an A-antimagic labeling and that G is an A-antimagic graph. The
integer-antimagic spectrum of a graph G is the set IAM(G) ={k | G is Zk-antimagic and k ≥ 2}.
Clearly IAM(G) ⊆ {k | k ≥ |V(G)|}. In this paper, we will use the term ‘labeling’ instead of edge labeling. Notation and concepts not defined here are referred to the book [1].
The concept of theA-antimagicness property for a graph G(introduced in [2]) naturally
arises as a variation of the A-magic labeling problem (where the induced vertex labeling
is a constant map) (for example, see [4–6, 8–10, 13]). The concept of ring-magic is another variation of group-magic (see [7]). It is also a variation of the anti-magic labeling problem (for example, see [16]) and edge-graceful labeling problem (for example, see [17]). The integer-antimagic spectra of some famous classes of graphs were determined [2, 3, 11, 12, 14, 15].
The following result is obvious.
Lemma 1.1 ( [2]). Let G = (V, E) be a simple graph and A be any abelian group. Let
f :E →A∗ Then ∑
v∈V
f+(v) = ∑
v∈V
∑
vu∈E
f(vu) = 2∑
e∈E
f(e). (1.1)
By using (1.1) and lettingA=Z4m+2 for some m≥0, we have
Lemma 1.2 ( [2, Lemma 1]). For m≥1, a graph of order 4m+ 2 is not Z4m+2-antimagic. For integersa≤b, let [a, b] denote the set of integers froma tob, inclusive. We shall still call it an interval.
Proposition 1.3. All elements in [a, b] are distinct after taking modulok for k≥b−a+ 1.
Let N be the set of all positive integers. Let f : E(G) → N be a labeling of a graph G and f+ be its induced vertex labeling. Denote the range of f+ as
If(G) = {f+(v)|v ∈V(G)}.
SupposeM is the maximum value (edge label) off. Sometimes we will refer to the range
of f+ and the maximum edge label of f in the same concise notation by
If(G) = {f+(v)|v ∈V(G)}▹(M).
Lemma 1.4 ( [14, Corollary 2.6]). For n ≥1, there is a labeling f for each of the following cycles such that If(C4n−1) = [3,4n+ 1]▹(2n+ 1), If(C4n) = [3,4n+ 2]▹(2n+ 1), If(C4n+1) = [2,4n+ 2]▹(2n+ 1) and If(C4n+2) = [3,4n+ 5]\ {4n+ 2}▹(2n+ 3).
Case 1. p= 4n, n≥1:
f(ei) =
{
i if 1≤i≤2n;
3 + 2(2n− ⌈2i⌉) if 2n+ 1≤i≤4n. Case 2. p= 4n+ 1, n ≥1:
f(ei) =
{
i if 1≤i≤2n;
3 + 2(2n− ⌈2i⌉) if 2n+ 1≤i≤4n+ 1. Case 3. p= 4n+ 2, n ≥1:
f(ei) =
{
i if 1≤i≤2n+ 3;
3 + 2(2n− ⌈i−22⌉) if 2n+ 4≤i≤4n+ 2. Case 4. p= 4n−1, n≥2:
f(ei) =
{
i if 1≤i≤2n+ 1;
3 + 2(2n− ⌈i+12 ⌉) if 2n+ 2≤i≤4n−1.
Case 5. p= 3: We label the edges of C3 by 1,2 and 3. HenceIf(C3) = [3,5].
By using each labeling defined above, we have the lemma.
ForS ⊂Z and a∈Z, we define the set a+S={a+s|s∈S}.
Lemma 1.5 ( [12, Lemma 3.1]). For n≥3, suppose g :E(Cn)→Z is a labeling and c∈Z.
Let h =g+c. Then Ih(Cn) = 2c+Ig(Cn).
Corollary 1.6. LetG be a disjoint union of cycles. Supposeg :E(G)→Zis a labeling and
c∈Z. Let h=g+c. Then Ih(G) = 2c+Ig(G).
Lemma 1.7. For n ≥ 1, there is a labeling h for C4n+2 such that Ih(C4n+2) = [2,4n+ 4]\ {4n+ 1}▹(2n+ 3).
Proof. Consider the labeling f for C4n+2 defined in Lemma 1.4. We reduce the label each edge e2j by 1. Let the resulting labeling be h. Then Ih(C4n+2) = [2,4n+ 4] \ {4n + 1}.
Clearly the maximum edge label is still 2n+ 3.
2
Disjoint unions of cycles of orders congruent to 0 modulo 4
LetG and H be two disjoint graphs. Let G+H be the disjoint union of G and H. The
disjoint union of n copies of a graph G is denoted bynG.
Suppose g and h are labelings of graphs G and H, respectively. The labeling g ∪h of
G+H is the labeling obtained by combining g and h. Applying Corollary 1.6 we have
Lemma 2.1. Let G be an even order disjoint union of cycles and let H be another disjoint union of cycles. Suppose there are labelings g and h of G andH such that Ig(G) = [3,|G|+
2]▹(m) and Ih(H) = [3,|H|+ 2]▹(M), where m ≤ |G|/2. Let f =g∪(h+|G|/2). Then
If(G+H) = [3,|G+H|+ 2]▹(M+|G|/2).
Lemma 2.2. Let m1, . . . , mα be positive integers. There is a labeling f of G= α
∑
i=1
C4mi such
Proof. There are labelingsfi ofC4mi such thatIfi(C4mi) = [3,4mi+2], 1≤i≤α. Applying
Lemma 2.1 repeatedly, we get the lemma.
3
Disjoint unions of cycles of orders congruent to 2 modulo 4
Let f be a labeling of a path P = x1x2· · ·xnxn+1, where n ≥ 1. For convenience, we let f(xixi+1) = ei (1 ≤ i ≤ n), and f+(xj) = aj (1 ≤ j ≤ n). Note that, in this case
f(x1x2) = a1 and f+(xn+1) = en. Suppose we know all the values of aj. Clearly, ei are
determined uniquely. Let us present this result precisely:
Since we have the relations aj = ej +ej−1, 2 ≤ j ≤ n,
i
∑
j=2
(−1)ja
j = a1 + (−1)iei, where
2≤i≤n. Thus
ei = i
∑
j=1
(−1)j+iaj. (3.1)
Note that
e2k = k
∑
j=1
(a2j −a2j−1)
e2k+1 =a1− 2∑k+1
j=2
(−1)jaj =a1+
k
∑
j=1
(a2j+1−a2j).
(3.2)
Lemma 3.1. Suppose n ≥ m. There are labelings η2 and η3 of G = C4m+2+C4n+2 such
that Iη2(G) = [2,|G|+ 1] and Iη3(G) = [3,|G|+ 2] with maximum edge label |G|/2 + 1. Proof. Let C4m+2 = u1u2· · ·u4m+2u1 and C4n+2 = v1v2· · ·v4n+2v1. We shall label C4m+2 first.
We divide the cycle C4m+2 into two (edge-disjoint) paths P =u1u2· · ·u2m+1u2m+2 and P′ =u1u4m+2u4m+1· · ·u2m+3u2m+2 of length 2m+ 1.
Firstly, we want to find a labelingϕforP satisfying thatϕ+(u
1) = 1 andϕ+(ui) = 4(i−1)
for 2≤i ≤2m+ 1; and a labeling ψ for P′ satisfying that ψ+(u
1) = 1,ψ+(u4m+2) = 3 and ψ+(u
4m+4−i) = 4i−3 for 3≤ i≤ 2m+ 1. By the discussion above, we know that ϕ and ψ
exist. From (3.2) we have
ϕ(u2m+1u2m+2) = ϕ+(u1) +
m
∑
j=1
4 = 1 + 4m,
ψ(u2m+3u2m+2) = ψ+(u1) + (ψ+(u4m+1)−ψ+(u4m+2)) +
m
∑
j=2 4
= 1 + 6 + 4(m−1) = 3 + 4m.
Let f = ϕ∪ψ. Thus, f+(u
Now, we keep the labeling above forC4m+2 and try to label the cycleC4n+2.
Case 1: m=n. Similarly, we divide the cycleC4n+2 into two pathsQ=v1v2· · ·v2n+1v2n+2 and Q′ =v1v4n+2v4n+1· · ·v2n+3v2n+2 of length 2n+ 1. And then, we want to find a labeling ϕ for Q satisfying that ϕ+(v
1) = 2 and ϕ+(vi) = 4i−2 for 2≤i≤2n+ 1; and a labeling ψ
for Q′ satisfying that ψ+(v
1) = 3 and ψ+(u4n+4−i) = 4i−1 for 2 ≤i≤2n+ 1. Now
ϕ(v2n+1v2n+2) = ϕ+(v1) +
n
∑
j=1
4 = 2 + 4n,
ψ(v2n+3v2n+2) = ψ+(v1) +
n
∑
j=1
4 = 3 + 4n.
Letg =ϕ∪ψ. We have g+(v
1) = 5, g+(v2n+2) = 8n+ 5, all edge labels are positive and the maximum label is 4n+ 3 underg. Moreover,Ig(C4n+2) = {5} ∪(4[1,2n] + 2)∪(4[1,2n] + 3). Let η2 =f ∪g. Here Iη2(C4m+2+C4n+2) = [2,4m+ 4n+ 5]▹(4m+ 3).
Case 2: m < n. Let n=m+k, where k ≥1. For convenience, we rewrite
C4n+2 =v1· · ·v2m+1y1· · ·y2kv2n+2z1· · ·z2kv4n−2m+3v4n−2m+4· · ·v4n+1v4n+2v1. Now we divide C4n+1 intoR =v1· · ·v2m+1y1· · ·y2kv2n+2 and R′ =v1v4n+2v4n+1· · ·v4n−2m+3z2k· · ·z1v2n+2.
We find a labelingϕ for R satisfying that ϕ+(v
1) = 2, ϕ+(vi) = 4i−2 for 2≤i≤2m+ 1
and ϕ+(y
i) = 8m+ 3 + 2ifor 1≤i≤2k; and a labelingψ forR′ satisfying thatψ+(v1) = 3, ψ+(u
4n+4−i) = 4i−1 for 2≤i≤2m+ 1 andψ+(z2k−i+1) = 8m+ 4 + 2ifor 1≤i≤2k. Now
ϕ(y2kv2n+2) =ϕ+(v1) +
m
∑
j=1 4 +
k
∑
j=1
2 = 2 + 4m+ 2k,
ψ(z1v2n+2) =ψ+(v1) +
m
∑
j=1 4 +
k
∑
j=1
2 = 3 + 4m+ 2k.
Letg =ϕ∪ψ. Since the required vertex label set for{vi |2≤i≤2m+ 1,4n−2m+ 3≤
i≤4n+ 2} is the same as Case 1 and g+(v1) = 5, [2,8m+ 4]⊂If∪g(C4m+2+C4n+2). From the requirement of vertices yj’s andzj’s, we get that [8m+ 5,4m+ 4n+ 4]⊂If∪g(C4m+2+ C4n+2). Finally we have g+(v2n+1) = 8m+ 4k+ 5 = 4m+ 4n+ 5. Let η2 = f ∪g. Here Iη2(C4m+2 +C4n+2) = [2,4m+ 4n+ 5]. Moreover, one may check that all edge labels are
positive and the maximum label is 4m+ 2k+ 3 = 2m+ 2n+ 3 under η2.
We define a new labeling η3 from η2 by increasing the label of each edge uiui+1 and vjvj+1 for odd i (1 ≤ i ≤ 4m+ 1) and odd j (1 ≤ j ≤ 4n + 1) by 1. Hence, Iη3(G) =
[3,|G|+ 2]▹(2m+ 2n+ 3).
Example 3.1. Here is a labeling η2 forC10+C10= 2C10 such that Iη2(2C10) = [2,21].
u10
u2
u1 u6
12 16
8 4
2
17 13 9
3
20
1
1
3 5 7
9
11
2 7 6
v10
v2
v1 v6
6
5
19 15
21
2
3
4 6 8
10
11
10 14 18
7 4 11 7 8
v2
v1 v10
v18
v5
v15
y1 y2 y3 y4
z1
z2
z3 z4
v3 v4
v16 v17 6 5 19 15 2 3 4
10 14 18
7 11
21 23 25
10 6 8 13 11 11 8 7 4
11 12 13
14
15
13
22 24 26 28
29 27
Now, one can see that Iη2(C10+C18) = [2,29].
Example 3.2. Here is a labeling η3 forC10+C10= 2C10 such that Iη3(2C10) = [3,22].
u10
u2
u1 u6
13 17 9 3 18 14 2 1
3 6 7
10
11
3 10 7 7
5
4
21
v10
v2
v1 v6
6
20 22
3
4 7 8
11
11
11 15 19
8 5 7 9
3
7
12 16
For C10+C18, we keep the labeling for the first C10 in 2C10 and the labeling for C18 is:
v2
v1 v10
v18
v5
v15
y1 y2 y3 y4
z1
z2
z3
z4
v3 v4
v16 v17 7 6 20 16 3 3 4
11 15 19
8 12
22 24 26
11 7 8 14 12 11 9 7 5
11 13 13
15
15
13
23 25 27 29
30 28
Now, one can see that Iη3(C10+C18) = [3,30].
4
Disjoint unions of cycles of orders congruent to 1 modulo 4
Lemma 4.1. There is a labeling fm for C4m+1 such that If(C4m+1) = [3,4m+ 4]\ {4m+ 1}▹(2m+ 2).
Proof. LetC4m+1 =u1u2· · ·u4m+1u4m+2, where u4m+2 =u1. Define
fm(uiui+1) =
1 + 2⌊2i⌋, if 1≤i≤2m;
2m+ 2, if i= 2m+ 1,2n+ 2; 4m+ 3−i, if 2m+ 3≤i≤4n+ 1.
We obtain that
fm+(ui) =
3, i= 1;
2i, if 2≤i≤2m;
4m+ 3, if i= 2m+ 1;
4m+ 4, if i= 2m+ 2;
4m+ 2, i= 2n+ 3;
8m+ 7−2i, if 2m+ 4 ≤i≤4m+ 1.
Lemma 4.2. There is a labelingfn forC4n+1 such thatIfn(C4n+1) ={3}∪[7,4n+ 7]\{4n+ 4}▹(2n+ 5).
Proof. LetC4n+1 =w1w2· · ·w4n+1w4n+2, wherew4n+2 =w1. Define
fn(wiwi+1) =
i, if i is odd and 1≤i≤2n−1;
4 +i, if i is even and 2≤i≤2n;
2n+ 2, if i= 2n+ 1;
2n+ 5, if i= 2n+ 2;
4n+ 3−i, if i is odd and 2n+ 3≤i≤4n+ 1; 4n+ 6−i, if i is even and 2n+ 4≤i≤4n.
We obtain that
fn+(vi) =
3, i= 1;
3 + 2i, if 2≤i≤2n;
4n+ 6, if i= 2n+ 1;
4n+ 7, if i= 2n+ 2;
4n+ 5, if i= 2n+ 3;
8n+ 10−2i, if 2n+ 4 ≤i≤4n+ 1.
Hence Ifn(C4n+1) ={3} ∪[7,4n+ 7]\ {4n+ 4}▹(2n+ 5).
Lemma 4.3. There is a labelingfsforC4s+1 such that Ifs(C4s+1) ={2}∪[6,4s+5]▹(2s+4). Proof. LetC4s+1 =v1v2· · ·v4s+1v4s+2, wherev4s+2 =v1. Define
fs(vivi+1) =
i, if i is odd and 1≤i≤2s+ 1;
3 +i, if i is even and 2≤i≤2s;
4s+ 2−i, if i is odd and 2s+ 3≤i≤4s+ 1; 4s+ 6−i, if i is even and 2s+ 2≤i≤4s.
We obtain that
fs+(vi) =
2, i= 1;
2 + 2i, if 2≤i≤2s+ 1;
4s+ 5, if i= 2s+ 2;
8s+ 9−2i, if 2s+ 3≤i≤4s+ 1.
Hence Ifs(C4s+1) ={2} ∪[6,4s+ 5]▹(2s+ 4).
Lemma 4.4. Letm, n, s, t be positive integers. There is a labelingη of G=C4m+1+C4n+1+ C4s+1+C4t+1 such that Iη(G) = [3,|G|+ 2]▹(|G|/2 + 1).
Proof. From Lemmas 4.1, 4.2, 4.3 and 1.4, we have a labelings fm, fn, fs and ft such that
Ifm(C4m+1) = [3,4m+ 4]\ {4m+ 1}, Ifn(C4n+1) ={3} ∪[7,4n+ 7]\ {4n+ 4}, Ifs(C4s+1) = {2} ∪[6,4s+ 5] and Ift(C4t+1) = [2,4t+ 2].
Letgn=fn+ (2m−1), gs =fs+ (2m+ 2n) and gt =ft+ (2m+ 2n+ 2s+ 2). Combining
fm, gn, gs and gt as a labeling η for C4m+1 +C4n+1+C4s+1+C4t+1 we have the required
5
Disjoint unions of cycles of orders congruent to 3 modulo 4
It is known that C3 is not Z3-antimagic [2, Theorem 3]. Following we will define some labelings for disjoint union of cycles whose orders are congruence 3 modulo 4. In particular, there are some labelings of disjoint unions of 3-cycles that are useful, even though they are not antimagic labelings.
Lemma 5.1. There are labelings g and h of C4n−1 such that Ig(C4n−1) = {3} ∪[7,4n+ 4]▹ (2n+ 4) and Ih(C4n−1) = [2,4n+ 1]\ {4n−2}▹(2n+ 1).
Proof. LetC4n−1 =u1u2· · ·u4n−1u4n, where u4n =u1. Define
g(uiui+1) =
i if i is odd and 1≤i≤2n−1;
4 +i if i is even and 2≤i≤2n;
4n+ 1−i if i is odd and 2n+ 1≤i≤4n−1; 4n+ 4−i if i is even and 2n+ 2 ≤i≤4n−2.
We obtain that
g+(ui) =
3, i= 1;
3 + 2i, if 2≤i≤2n;
4n+ 4, if i= 2n+ 1;
8n+ 6−2i, if 2n+ 2≤i≤4n−1.
Hence Ig(C4n−1) ={3} ∪[7,4n+ 4].
When n≥2. Define
h(uiui+1) =
i if i is odd and 1≤i≤2n−3;
1 +i if i is even and 2≤i≤2n−2; 1 +i if 2n−1≤i≤2n;
4n−i if 2n+ 1 ≤i≤4n−1.
We obtain that
h+(ui) =
2, i= 1;
2i, if 2≤i≤2n−2;
4n−1, if i= 2n−1;
4n+ 1, if i= 2n;
4n, if i= 2n+ 1;
8n+ 1−2i, if 2n+ 2 ≤i≤4n−1.
Hence Ih(C4n−1) = [2,4n+ 1]\ {4n−2}. Note that, when n= 1, h=f, wheref is defined
in Lemma 1.4. Hence Ih(C3) = [3,5]. So it still holds.
Proof. Let f be the labeling defined in Lemma 1.4 and h be the labeling defined in Lemma 5.1. Let f3 = f∪(h+ 2m). Here If3(G) = [3,4m+ 4n+ 1]\ {4m+ 4n−2} with
maximum edge label 2m+ 2n+ 1.
Lemma 5.3. There is a labeling ℓ of C4n−1 such that Iℓ(C4n−1) = {2} ∪[6,4n+ 4]\ {4n+ 1}▹(2n+ 4).
Proof. Keeping the notation as Lemma 5.1, we define
ℓ(uiui+1) =
i if i is odd and 1≤i≤2n−1;
3 +i if i is even and 2≤i≤2n−2;
2n if i= 2n+ 1;
4n−i if i is odd and 2n+ 3≤i≤4n−1;
4n+ 4−i if i is even and 2n≤i≤4n−2.
We obtain that
ℓ+(ui) =
2 if i= 1;
2 + 2i, if 2≤i≤2n−1;
4n+ 3, if i= 2n;
4n+ 4, if i= 2n+ 1;
4n+ 2, if i= 2n+ 2;
8n+ 5−2i, if 2n+ 3≤i≤4n−1.
Hence Iℓ(C4n−1) = {2} ∪[6,4n+ 4]\ {4n+ 1}.
Corollary 5.4. Let G = C4m−1 +C4n−1, where m ≤ n and n ≥ 2. There is a labeling f6
such that If6(G) = {2} ∪[6,|G|+ 4]▹(|G|/2 + 4).
Proof. Let g and ℓ be labelings defined in Lemmas 5.1 and 5.3, respectively. Hence Ig(C4m−1) = {3} ∪[7,4m + 4] and Iℓ(C4n−1) = {2} ∪[6,4n + 4]\ {4n + 1}. Let f6 =
ℓ∪(g+ (2n−1)). Here If6(G) = {2} ∪[6,4m+ 4n+ 2]▹(2m+ 2n+ 3).
Lemma 5.5. There is a labeling f8 of 2C3 such that If8(2C3) ={4} ∪[8,12]▹(8).
Proof. Let f8 be the labeling of 2C3 by labeling the edges of the first C3 by 1,3,8 and the
edges of the second C3 by 3,5,7. Then If8(2C3) ={4} ∪[8,12].
Note thatf8 is neither a Z7-antimagic labeling nor aZ8-antimagic labeling, since 0 would be an edge label in each.
Corollary 5.6. Let G =C4m−1 +C4n−1+C4s−1, where m ≤ n ≤ s. There is a labeling η1
such that Iη1(G) = [2,|G|+ 1] when s ≥2 and Iη1(3C3) = [4,12]. The maximum edge label
is (|G|+ 1)/2 + 2.
Proof. Let h be the labeling of C4m−1 defined in Lemma 5.1 and f6 be the labeling of C4n−1+C4s−1 defined in Lemma 5.5.
When s≥2. Letη1 =h∪(f6+ (2m−2)). Hence Iη1(G) = [2,4m+ 4n+ 4s−2]▹(2m+
When s = 1. In this case G = 3C3. We label the edges of these three C3 by 1,4,3; 2,6,4; 4,7,5 accordingly. Denote this labeling byη1. ThenIη1(3C3) = [4,12]. Note that the
maximum edge label is still 2m+ 2n+ 2s+ 1.
Lemma 5.7. Let G = C4m−1 +C4n−1 +C4s−1 +C4t−1, where m ≤ n ≤ s ≤ t. There
is a labeling ϕ such that Iϕ(G) = [3,|G|+ 2] ▹ (|G|/2 + 3) when t ≥ 2 and Iϕ(4C3) = [3,14]▹(|4C3|/2 + 2).
Proof. Consider labelingsf forC4m−1 defined in Lemma 1.4 andη1forC4n−1+C4s−1+C4t−1 defined in Corollary 5.6.
When t≥2. Letϕ =f ∪(η1+ 2m). Here Iϕ(G) = [3,|G|+ 2]▹(|G|/2 + 3). When t= 1.
Let ϕ=f∪(η1+ 1). HereIϕ(G) = [3,14]▹(8).
6
Main result
Before considering the general case, we introduce another lemma first.
Lemma 6.1. There is a labeling σ3 for G=C4m+1+C4n−1 such that Iσ3(G) = [3,|G|+ 2]▹
(|G|/2 + 1).
Proof. From Lemma 1.4 there are labelingsf andgsuch thatIf(C4m+1) = [2,4m+2]▹(2m+ 1) andIg(C4n−1) = [3,4n+1]▹(2n+1). Letσ3 =g∪(f+2n). HereIσ3 = [3,|G|+2]▹(|G|/2+1).
Now consider a disjoint union of cycles
G=
α
∑
i=1
C4mi + β
∑
j=1
C4nj+1+
γ
∑
k=1
C4sk+2+ δ
∑
l=1
C4tl−1,
whereα, β, γ, δ≥0 and mi, nj, sk, tl are positive. Note that, as usual, a summation is empty
if its upper limit is less than its lower limit. Since the integer-antimagic spectra of single cycle has been determined [2], we may assume that G contains at least two cycles.
Supposeβ = 4β′+β0,γ = 2γ′+γ0, andδ= 4δ′+δ0, where 0≤β0, δ0 ≤3 and 0≤γ0 ≤1.
When β0 ≥ δ0 ≥ 1, let G0 =
δ
∑
l=4δ′+1
(C4nl+1 +C4tl−1). Similarly, when δ0 ≥ β0 ≥ 1, let
G0 =
β
∑
j=4β′+1
(C4nj+1+C4tj−1).
LetG1 =
α
∑
i=1
C4mi+
4β′
∑
j=1
C4nj+1+ 2γ′
∑
k=1
C4sk+2+ 4δ′
∑
l=1
C4tl−1+G0. We may rewrite the remaining
part of G as G−G1 =
β1
∑
j=1
C4pj+1+
γ0
∑
k=1
C4rk+2 +
δ1
∑
l=1
C4ql−1, where 0 ≤γ0 ≤ 1, 0≤β1, δ1 ≤3
and at least one of β1 and δ1 is zero.
From Lemmas 2.2, 4.4, 3.1, 5.7 and 6.1 and applying Lemma 2.1 repeatedly, there are
labelings for
α
∑
i=1 C4mi,
4β′
∑
j=1
C4nj+1, 2γ′
∑
k=1
C4sk+1, 4δ′
∑
l=1
C4tl−1 and G0 such that the range of each
Suppose G1 exists. Applying Lemma 2.1 on G1 repeatedly, we have a labeling θ of G1 such that Iθ(G1) = [3,|G1|+ 2]▹(|G1|/2 + 1).
Following we will deal with the remaining partG−G1 if any. So we assume that at least one of β1, γ0, δ1 is nonzero. Let
H1 =
β1
∑
j=1
C4pj+1+
δ1
∑
l=1
C4ql−1, if γ0 = 0,
H2 =H1+C4r+2 for some r≥1, if γ0 = 1,
where 0 ≤β1, δ1 ≤3 and at least one of β1 and δ1 is zero. Case 1: Consider the case whenγ0 = 0, that isG=G1+H1.
1-1: Suppose β1 = 0.
When δ1 = 1. By using the labeling f in Lemma 1.4 we have If(H1) = [3,|H1|+ 2]. Let σ=θ∪(f +|G1|/2). Here Iσ(G) = [3,|G|+ 2]▹(|G|+ 1)/2 + 1).
When δ1 = 2. By using the labeling f3 in Corollary 5.2, we have If3(H1) = [3,|H1|+
3]\ {|H1|}▹(|H1|/2 + 2). Letσ =θ∪(f3+|G1|/2). HenceIσ(G) = [3,|G|+ 3]\ {|G|+
2}▹(|G|/2 + 2).
When δ1 = 3. Choose the labeling η1 in Corollary 5.6. Let σ = η1 ∪(θ+x), where x is (|H1| −1)/2 or (|H1|+ 1)/2 for H1 ̸= 3C3 or H1 = 3C3, respectively. Here Iσ(G) = [2,|G|+ 1] or [4,|G|+ 3] and the maximum label at most (|G|+ 1)/2 + 1.
1-2: Suppose δ1 = 0.
When β1 = 1. Let f be the labeling in Lemma 1.4. Let σ = f ∪(θ+ (|H1| −1)/2). Hence, Iσ(G) = [2,|G|+ 1]▹((|G| −1)/2).
When β1 = 2. Combining the labeling in Lemma 4.1 with a translation of the labeling
in Lemma 4.2 we have a labeling f′ for H1 such that If′(H1) = [3,|H1|+ 3]\ {|H1|}▹ (|H1|/2 + 3). Let σ =θ∪(f′+|G1|/2). Hence, Iσ(G) = [3,|G|+ 3]\ {|G|}▹(|G|/2 + 3).
When β1 = 3. Using the labelings defined in Lemma 4.1, Lemma 4.2 and Lemma 4.3,
we have a labeling f′ for H1 such that If′(H1) = [3,|H1|+ 2]▹((|H1|+ 1)/2 + 1). Let σ =θ∪(f′ +|G1|/2). Hence, Iσ(G) = [3,|G|+ 2]▹((|G|+ 1)/2 + 1).
Case 2: Consider the case when γ0 = 1, that is G=G1 +H1+C4r+2. Here H1 may not exist.
2-1: When (β1, δ1) = (0,0). In this case, G = G1 +C4r+2. We have the labeling θ for G1 such thatIθ(G1) = [3,|G1|+ 2]▹(|G1|/2 + 1). From Lemma 1.4 there is a labelingf for C4r+2 such that If(C4r+2) = [3,4r+ 5]\ {4r+ 2}▹(2r+ 3). Let ρ=θ∪(f+|G1|/2). Then Iρ(G) = [3,|G|+ 3]\ {|G|}▹(|G|/2 + 2).
2-2: When (β1, δ1) = (0,2). In this case, G = G1+C4r+2 +C4q1−1+C4q2−1. Let ρ be the
labeling in Case 2-1. Let L1 =C4q1−1 +C4q2−1.
Suppose L1 ̸= 2C3. From Corollary 5.4 we have a labeling f6 for L1 = such that If6(L1) ={2} ∪[6,|L1|+ 4]▹(|L1|/2 + 4). Letρ1 =ρ∪(f6+|G1+C4r+2|/2−1). Here
Suppose L1 = 2C3. From Lemma 5.5 we have a labelingf8 for L1 such that If8(L1) = {4} ∪[8,12]▹(8). Let ρ1 =ρ∪(f8+|G1+C4r+2|/2−2). HereIρ1(G) = [3,|G|+ 2]▹
(|G|/2 + 3).
2-3: When (β1, δ1) = (2,0). In this case,G=G1+C4r+2+C4p1+1+C4p2+1. LetL2 =C4p1+1+
C4p2+1. Combining the labeling in Lemma 4.3 with a translation of the labeling in
Lemma 1.4, we have a labelingfforL2such thatIf(L2) = {2}∪[6,|L2|+4]▹(|L2|/2+4). Similar to Case 2-2 we have a labelingρ2forGsuch thatIρ2(G) = [3,|G|+2]▹(|G|/2+3).
2-4: When (β1, δ1) = (0,1) or (3,0). Let σ be the labeling of G1 +H1 in Case 1 and h be the labeling in Lemma 1.7. Let ρ3 =σ ∪(h+ (|G1 +H1|+ 1)/2). Then Iρ3(G) =
[3,|G|+ 3]\ {|G|}▹((|G|+ 1)/2 + 2).
2-5: When (β1, δ1) = (0,3) or (1,0). Let σ be the labeling of G1 +H1 in Case 1 and f be the labeling in Lemma 1.4.
Suppose H1 ̸= 3C3. Letρ4 =σ∪(f + (|G1+H1| −1)/2). Then Iρ(G) = [2,|G|+ 2]\
{|G| −1}▹((|G|+ 1)/2 + 1).
Suppose H1 = 3C3. Let ρ4 =σ∪(f+ (|G1+H1|+ 1)/2). ThenIρ(G) = [4,|G|+ 4]\
{|G|+ 1}▹((|G|+ 1)/2 + 2).
SupposeG1 does not exist. Then the proof is similar and simpler than the case whenG1 exists. So we omit here.
From the discussion in this paper, we can see that all the edge labels are positive and at most ⌊|G|/2⌋+ 3, where G is the considered graph. If |G| ≥ 7, then ⌊|G|/2⌋+ 3 < |G|. So, for these cases, all labelings are proper when we take modulo k, where k ≥ |G|. If 3 ≤ |G| ≤ 6 and G contains at least two cycles, then G must be 2C3. Corollary 5.2 shows that If3(2C3) = [3,9]\ {6}▹(5). Thus, the labeling is proper.
Applying Proposition 1.3 we get
Theorem 6.2. SupposeG is a disjoint union of cycles of order p. Then
IAM(G) =
[4,∞) if p= 3,
[p,∞) if p̸≡2 (mod 4) and p̸= 3,
[p+ 1,∞) if p≡2 (mod 4).
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